By this time you should know a great deal about DC circuits. In particular,
you should be familiar with the following.
Resistance
Ohm's Law
Series & Parallel
Combinations of Resistors
Kirchhoff's Current Law
(KCL)
Kirchhoff's Voltage Law
(KVL)
The Method of Node Volages
The Method of Mesh Equations
Thevinin and Norton Equivalent
Circuits
When
you learned these topics you might have assumed that what you were learning
was something that only applied to DC circuits. In a way you were
right, but in a way you were wrong. While the methods you learned
only apply to DC circuits, the methods you learned can also be generalized
to AC circuits. It will take a little work, but every one of the
analysis tools listed above has a related - and very similar - AC version.
If you want to generalize your tool set (all of the tools listed above)
you will first need to learn about phasors.
Here is a short lesson on phasors and an introduction to using phasors.
A sinusoidal signal -
voltage or current or any other physical variable - can be represented
by a phasor .
Call that phasor V.
If the sinusoidal signal
is given by:
v(t) = V_{max}
cos(wt
+ f)
NOTE: We could also
use: v(t) = V_{max} sin(wt
+ f).
What matters is the relative phase between signals.
Then, the signal can be
represented by a phasor:
v(t) <=> V_{max}/f
The little symbol "<=>"
means that the phasor - the expression on the right - represents a voltage
function of time.
The voltage function of
time is sinusoidal with an amplitude of V_{max}, and a phase
angle of f.
and we would write an
expression for the phasor
V:
V = V_{max}/f
Phasors in linear circuits
are related if all of the signals in the circuit are at the same frequency.
If you have signals that are different frequencies you can't use phasors
to figure out how signals add, etc.
Important
Properties of Phasors - Adding Phasors
In this section we will examine some simple properties of phasors.
We start with one that may or may not be obvious.
Adding phasors is equivalent
to adding the corresponding time function for each phasor.
To
see how this works out, consider the sum of two voltage phasors
V_{sum}=
V_{1}
+ V_{2} = V_{1}/f1
+ V_{2}/f2
This sum corresponds to the sum of the two
voltages (as might occur when you write KVL). The actual sum would
be written:
v_{sum}(t)
= V_{sum} cos(wt
+ f_{sum})
= V_{1}
cos(wt
+ f1)
+ V_{2} cos(wt
+ f2)
The problem is to be able to use the phasor
method to do the addition, then interpret that in terms of the time functions.
You and interpret the problem in at least two different ways.
One way to interpret this
is in terms of complex numbers. The sum voltage - represented as
a phasor - is equal to the sum of the other two voltages - represented
as phasors.
V_{sum}
= V_{1} + V_{2}= V_{1}/f1
+ V_{2}/f2
Each phasor can be represented
by
a complex number. Break each phasor into real and imaginary
parts.
V_{1}
= V_{1}cos(f1)
+ jV_{1}sin(f1)
V_{2}
= V_{2}cos(f2)
+ jV_{2}sin(f2)
So, the sum of the two
phasors can be computed by adding the real and the imaginary parts separately,
giving:
Then, we can note the
the real part and the imaginary part are the real and imaginary parts of
the sum.
= V_{sum,real}
+ jV_{sum,imaginary}
And, the phasor for the
sum voltage can also be represented with a magnitude-angle representation.
= V_{sum}cos(f_{sum})
+ jV_{sum}sin(f_{sum})
There
is, of course, also a geometrical interpretation whenever you deal with
complex numbers or variables. If we have two phasors that we are
adding, we visualize the situation as shown below.
V_{sum}
= V_{1} + V_{2}
We can add the two phasors any way possible. That includes doing
it graphically by hand, breaking the phasors
into components and summing the real and imaginary
components - as we did above - or any other way you can imagine
to sum two vector-like quantities.
Differentiation
of Time Functions Represented by Phasors
A second
operation that we need to perform often is differentiation
of a time function represented by a phasor. Consider a sinusoidal
time function, v(t):
v(t) = V_{max}
cos(wt
+ f)
with a phasor
V:
V = V_{max}/f
The derivative of the time function is given
by:
v_{d}(t)
= dv(t)/dt = - wV_{max}
sin(wt
+ f)
The phasor for the derivative signal is:
V_{d}
= - wV_{max}/f-90^{o}=
wV_{max}/f+90^{o}
(Note the minus sign on the first represention
where ninety degrees is subtracted has been eliminated when the angle is
changed to plus ninety degrees!)
We conclude the following.
Differentiating a sinusoidal
signal is equivalent to multiplication of the signal's phasor by w
and rotation of the phasor by 90^{o}.
There
is also an interpretation in terms of complex numbers. Represent
the phasor V in terms of complex numbers:
V = Vcos(f)
+ jVsin(f)
Then, consider the
phasor for the derivative:
V_{d}
= V_{d}cos(f)
+ jV_{d}sin(f)
But, we can also write
the phasor for the derivative from the time function for the derivative.
V_{d}
= wVcos(f+90^{o})
+ jwVsin(f+90^{o})
Now, work with this expression.
Take the ninety degree
terms out of the arguments since:
cos(f+90^{o})
= -sin(f),
and sin(f+90^{o})
= cos(f).
V_{d}
= - wVsin(f)
+ jwVcos(f)
and with some insight, we can note that this
is the same as (since j^{2} = -1):
V_{d}
= j^{2}wVsin(f)
+ jwVcos(f)
V_{d}
= j w(
jVsin(f)
+ Vcos(f))
= j wV
The conclusion:
Differentiating a sinusoidal
signal is equivalent to multiplication of the signal's phasor by jw.
There is a little side note here that is important. When you study
Laplace transforms you will find that differentiating a signal - any signal,
not just a sinusoid - is equivalent to multiplication of the signal's Laplace
transform by the Laplace transform variable, s. The similarity is
not an accident. The Laplace transform is a generalization of the
Fourier transform, and the Fourier transform is based on the idea that
a signal - almost any signal - can be represented as a sum of sinusoids.
In the case of the Fourier transform, that sum becomes an integral.
When you travel this road, you go through Fourier
Series, then get to Fourier transforms and finally you get to Laplace
transforms. Not everybody takes all of the steps, and some might
jump directly to Laplace transforms, but viewing the entire path can be
enlightening.
Example
E1
In a capacitor, the voltage and current are related by:
i_{c}(t)
= C dv_{c}(t)/dt
so the voltage phasor,
V_{c},
and the current phasor, I_{c},
are related by:
I_{c}
= jwCV_{c}
E2
Imagine that you have two voltage sources in series as shown below.
These two voltages
are both sinusoidal, but at different phases. You can see the voltages
in this simulator.
To see the cosine term,
type a "1" in the cosine term text-box and be sure that the cosine term
stays zero.
To see the sine term,
type a "1" in the sine term text-box, and set the cosine term to 0.